Question: Square $ABCD$ is inscribed in the region bound by the parabola $y = x^2 - 8x + 12$ and the $x$-axis, as shown below.  Find the area of square $ABCD.$

[asy]
unitsize(0.8 cm);

real parab (real x) {
  return(x^2 - 8*x + 12);
}

pair A, B, C, D;
real x = -1 + sqrt(5);

A = (4 - x,0);
B = (4 + x,0);
C = (4 + x,-2*x);
D = (4 - x,-2*x);

draw(graph(parab,1.5,6.5));
draw(A--D--C--B);
draw((1,0)--(7,0));

label("$A$", A, N);
label("$B$", B, N);
label("$C$", C, SE);
label("$D$", D, SW);
[/asy]
Explanation: Note that the axis of symmetry of the parabola is $x = \frac{-(-8)}{2\cdot1}=4.$

Let $2t$ be the side length of the square.  Then
\begin{align*}
A &= (4 - t, 0), \\
B &= (4 + t, 0), \\
C &= (4 + t, -2t), \\
D &= (4 - t, -2t).
\end{align*}But $C$ lies on the parabola $y = x^2 - 8x + 12 = (x - 4)^2 - 4,$ so
\[-2t = t^2 - 4.\]Then $t^2 + 2t - 4 = 0,$ so by the quadratic formula,
\[t = -1 \pm \sqrt{5}.\]Since $t$ is half a side length, it must be positive, and so $t = -1 + \sqrt{5}.$  Therefore, the area of the square is
\[(2t)^2 = (-2 + 2 \sqrt{5})^2 = \boxed{24 - 8 \sqrt{5}}.\]